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%高等代数（二）期末考试A
%{\LARGE\bf 上海立信会计金融学院期终考试卷 --- 试题纸} \hspace{0.3cm} {\Large \underline{ A }卷 }
{\Large\bf \H 上海立信会计金融学院期终考试卷 } \hspace{0.3cm} {\Large \underline{ A }卷 参考答案 }

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{\large \bf \H 2022 $\sim$ 2023 学年 第 二 学期 }

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{\large \bf \H \underline{ \emph{2022级数学与应用数学专业} } 《\underline{ \emph{高等代数(二)} }》 课程代码：\underline{ 160370410}  }


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%\author{2022级数学与应用数学、2022级应用统计学 }
%\author{学号 \underline{\hspace{4cm}} 姓名  \underline{\hspace{4cm}} }
%\title{高等代数二考试解答 }
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%\date{2023年5月31日}
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%\begin{abstract}
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%7.3. 
%7.4. 
%7.5. 

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\begin{enumerate}

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\item %第1题
设有二次型 $f(x_1,x_2,x_3)=5x_1^2+5x_2^2+cx_3^2-2x_1x_2+6x_1x_3-6x_2x_3$ 的秩为 2. %记 $X=\begin{bmatrix} x_1\\ x_2 \end{bmatrix}$.
\begin{enumerate}
\item  求 $c$;
\item  求 $f(x_1,x_2,x_3)$ 的矩阵;
\item  将 $f(x_1,x_2,x_3)$ 化为标准型;
\item  求该二次型的正惯性指数及符号差。
\end{enumerate}

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{\color{red} 解答：
\begin{enumerate}


\item  
由 $f(X)=X^tAX$ 可得对称矩阵 $A$, 即 
$$
5x_1^2+5x_2^2+cx_3^2-2x_1x_2+6x_1x_3-6x_2x_3 = X^tAX =
\begin{pmatrix}
x_1 & x_2 & x_3
\end{pmatrix}
\begin{pmatrix}
5 & -1 & 3 \\
-1 & 5 & -3 \\
3 & -3 & c \\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 
\end{pmatrix}. 
$$
%所以这个二次型对应的对称矩阵为
%$A=\begin{pmatrix}
%5 & -1 & 3 \\
%-1 & 5 & -3 \\
%3 & -3 & c \\
%\end{pmatrix}$, 
根据题意知矩阵 $A$ 的秩为2, 所以 $\det{A}=0$, 由此可得 $c=3$.

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  由上一小题可知 $c=3$, 所以该二次型的矩阵为
$
A=\begin{pmatrix}
5 & -1 & 3 \\
-1 & 5 & -3 \\
3 & -3 & 3 \\
\end{pmatrix}. 
$


\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

%(3) $A$ 不正定, 从而 $f$ 不是正定的.

\item  
方法一是先计算矩阵 $A$ 的特征值为 $4,9,0$. 所以标准型为 $f(y_1, y_2, y_3)=4y_1^2+9y_2^2$. 

方法二是使用配方法，
\begin{eqnarray*}
f(x_1,x_2,x_3) &=& 5x_1^2+5x_2^2+3x_3^2-2x_1x_2+6x_1x_3-6x_2x_3 \\
&=& 5(x_1 -\frac{1}{5}x_2 + \frac{3}{5}x_3)^2 + \frac{24}{5}x_2^2 - \frac{24}{5}x_2x_3 + \frac{6}{5}x_3^2 \\ 
&=& 5(x_1 -\frac{1}{5}x_2 + \frac{3}{5}x_3)^2 + \frac{24}{5} (x_2 - 2x_3)^2 
= 5y_1^2+\frac{24}{5}y_2^2. 
\end{eqnarray*}

方法三是使用合同变换，对矩阵 $A$ 依次进行相同的初等行变换与初等列变换，将矩阵 $A$ 化为对角阵
将第一行乘以 $\frac{1}{5}$ 加到第二行，将第一行乘以 $-\frac{3}{5}$ 加到第三行，
然后再将第一列乘以 $\frac{1}{5}$ 加到第二列，将第一列乘以 $-\frac{3}{5}$ 加到第三列，可得
$$
A=\begin{pmatrix}
5 & -1 & 3 \\
-1 & 5 & -3 \\
3 & -3 & 3 \\
\end{pmatrix}
\longrightarrow
\begin{pmatrix}
5 & -1 & 3 \\
0 & \frac{24}{5} & -\frac{12}{5} \\
0 & -\frac{12}{5} & \frac{6}{5} \\
\end{pmatrix}
\longrightarrow
\begin{pmatrix}
5 & 0 & 0 \\
0 & \frac{24}{5} & -\frac{12}{5} \\
0 & -\frac{12}{5} & \frac{6}{5} \\
\end{pmatrix}.
$$
接下来将第二行乘以 $\frac{1}{2}$ 加到第三行，将第二列乘以 $\frac{1}{2}$ 加到第三列，可得
$$
\begin{pmatrix}
5 & 0 & 0 \\
0 & \frac{24}{5} & -\frac{12}{5} \\
0 & -\frac{12}{5} & \frac{6}{5} \\
\end{pmatrix}
\longrightarrow
\begin{pmatrix}
5 & 0 & 0 \\
0 & \frac{24}{5} & -\frac{12}{5} \\
0 & 0 & 0 \\
\end{pmatrix}
\longrightarrow
\begin{pmatrix}
5 & 0 & 0 \\
0 & \frac{24}{5} & 0 \\
0 & 0 & 0 \\
\end{pmatrix}.
$$

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  由标准型可知，该二次型的正惯性指数为 2, 负惯性指数为0, 所以符号差为 2. 

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\end{enumerate}

}

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\item %第2题
设 $a$ 是一个实数，设 $f(x)=x^3+6x^2+3ax+8$. 讨论 $a$ 的值使得 $f(x)$ 有重因式。

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{\color{red} 解答：
导数多项式为 $f'(x)=3(x^2+4x+a)$. 将使用的定理是，多项式 $f(x)$ 有重因式当且仅当 $f(x)$ 与 $f'(x)$ 由公因式。
因此使用辗转相除法求公因式。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\begin{enumerate}
\item  当 $a=4$ 时，$(f(x),f'(x))=(x+2)^2$, 此时$x+2$ 是 $f(x)$ 的 3 重因式。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  当 $a=-5$ 时，$(f(x),f'(x))=x-1$, 此时 $x-1$ 是 $f(x)$ 的 2 重因式。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\end{enumerate}
}

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\item %第3题
向量空间 $\mathbb{R}^3$ 中的任意向量可以写作 $(x_1, x_2,x_3 )^T$, 设 $\mathbb{R}^3$ 上的一个变换 $\sigma$ 定义为 
$$\sigma \left(\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}}\end{array}\right)
=\left( \begin{array}{c}{x_{1}+2 x_{2}} \\ {x_{3}-x_{2}} \\ {x_{2}-x_{3}}\end{array}\right).$$

\begin{enumerate}
\item  证明:  ~$\sigma$ 为~$\mathbb{R}^3$ 上一个线性变换。
\item  在~$\mathbb{R}^3$ 上取一个基 $\epsilon_1=(1,0,0 )^T$, $\epsilon_2=(0,1,0 )^T$, $\epsilon_1=(0,0,1 )^T$.
求~$\sigma$ 在这个基~ $\{ \epsilon_1, \epsilon_2, \epsilon_3 \} $ 下的矩阵, 即求矩阵~$A$, 使得
$\sigma\left(\epsilon_{1}, \epsilon_{2}, \epsilon_{3}\right)=\left(\epsilon_{1}, \epsilon_{2}, \epsilon_{3}\right) A$. 
\item  在~$\mathbb{R}^3$ 上另取三个向量~ $\eta_1=(1,2,0 )^T$, $\eta_2=(1,0,-1 )^T$, $\eta_1=(0,-2,1 )^T$.
证明~ $\{ \eta_1, \eta_2, \eta_3 \} $ 仍为 ~$\mathbb{R}^3$ 的一个基，并求~$\sigma$ 在这个基~ $\{ \eta_1, \eta_2, \eta_3 \} $ 下的矩阵。

\end{enumerate}

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{\color{red} 解答：
\begin{enumerate}
\item  按照线性变换的定义，验证对任意 $k_1, k_2\in \mathbb{R}$, 以及任意 $\alpha, \beta\in \mathbb{R}^3$, 
都成立 $$\sigma(k_1\alpha+k_2\beta)=k_1\sigma(\alpha)+k_2\sigma(\beta). $$
另一种证明是将该变换写成 $\sigma(X)=AX$ 的形式，其中 $A$ 是一个 $3\times 3$ 的矩阵。然后引用结论，形如 $\sigma(X)=AX$ 的变换 $\sigma$ 都是线性变换。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  按照线性变换关于一个基的矩阵的定义，即 
$\left( \sigma(\epsilon_1), \sigma(\epsilon_2), \sigma(\epsilon_3) \right) 
= \left(\epsilon_1, \epsilon_2, \epsilon_3\right) A$, 计算左边可得
$$
\sigma(\epsilon_1) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},
\sigma(\epsilon_2) = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix},
\sigma(\epsilon_3) = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}. 
$$
因此所求矩阵为 
$A=\begin{pmatrix}
1 & 2 & 0 \\
0 & -1 & 1 \\
0 & 1 & -1 \\
\end{pmatrix}
$.

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  因为这三个列向量组成的矩阵 $C$ 的行列式为 
$ |C| = 
\begin{vmatrix}
1 & 1 & 0 \\
2 & 0 & -2 \\
0 & -1 & 1 \\
\end{vmatrix} \neq 0$, 
所以向量组 $\{\eta_1, \eta_2, \eta_3\}$ 线性无关，仍为 $\mathbb{R}^3$ 的一个基。 
设线性变换 $\sigma$ 关于 $\eta_1, \eta_2, \eta_3$ 的矩阵为 $B$, 则按照定义，有
 $$\left( \sigma(\eta_1), \sigma(\eta_2), \sigma(\eta_3) \right) 
= \left(\eta_1, \eta_2, \eta_3\right) B. $$
因为 $\left(\eta_1, \eta_2, \eta_3\right) = \left(\epsilon_1, \epsilon_2, \epsilon_3\right)\cdot C$, 
所以 
$B = C^{-1}AC = -\frac{1}{4}
\begin{pmatrix}
-12 &-3 & 11  \\
-8 &-1  &5  \\
-16& -5& 17  \\
\end{pmatrix}
$.

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\end{enumerate}

}

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\item %第4题
考虑实数域上的四维列向量空间 $V=\mathbb{R}^4$, 设 $\xi=(x_1,x_2,x_3,x_4)^T$ 与$\eta=(y_1,y_2,y_3,y_4)^T$ 是其中两个向量。
设 $V$ 中内积定义为 $\langle\xi,\eta\rangle=x_1y_1+x_2y_2+x_3y_3+x_4y_4$. 
设向量子空间 $W$ 为齐次线性方程组 
$$\left\{\begin{array}{l} x_1-2x_2+3x_3-4x_4=0\\ x_1+5x_2+3x_3+3x_4=0 \end{array}\right. $$ 的解空间。
\begin{enumerate}
\item  验证向量 $\alpha_1=(-3,0,1,0)^T$ 是 $W$ 中的一个向量。
\item  求 $W$ 中 与 $\alpha_1$ 的内积为零的所有向量组成的集合。
\item  求 $W$ 中的另一个向量 $\beta$, 使 $\alpha_1,\beta$ 成为 $W$ 的一个正交基。
\item  求 $W$ 的一个标准正交基。
\end{enumerate}


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{\color{red} 解答：
\begin{enumerate}
\item  将向量 $\alpha_1$ 代入齐次线性方程组可知等式成立。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  首先计算这个齐次线性方程组的基础解系为 
$$\alpha_1=(-3,0,1,0)^T, \beta=(2,-1,0,1)^T.$$ 
因此解空间为 $W=L(\alpha_1, \beta)$. 
令所求向量为 $\gamma= k\alpha_1+m\beta$, 
则 $\gamma = (-3k+2m,-m,k,m)^T$. 
根据题目条件 $(\gamma, \alpha_1)=0$ 可得 $(-3)(-3k+2m) +k =0$, 即 $10k=6m$. 
所以 $\gamma=m(1,-5,3,5)^T$, 其中 $m$ 为任意实数。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  由上一小题可得，所求向量为 $\beta=m(1,-5,3,5)^T$, 其中 $m$ 为任意非零实数。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  将 $\alpha_1$ 与 $\beta$ 单位化，即得标准正交基 
$\xi=\frac{1}{\sqrt{10}}(-3,0,1,0)^T, \eta= \frac{1}{2\sqrt{15}}(1,-5,3,5)^T$. 
因为 $W$ 的维数是2，所以存在无穷多个标准正交基。
%(5) $(1,0,0,0)^T, (0,1,0,0)^T$ (答案不唯一).

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\end{enumerate}

}

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\item %第5题
设 $A=\begin{pmatrix}
1 & 0 & 0 \\
2 & x & 2 \\
3 & 1 & 1 \\
\end{pmatrix}, 
B=\begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & y \\
\end{pmatrix}
$. 已知 $A$ 与 $B$ 相似。
\begin{enumerate}
\item  求 $x$ 和 $y$.
\item  求可逆阵 $P$ 使 $P^{-1}AP=B$.
\end{enumerate}

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{\color{red} 解答：
\begin{enumerate}
\item  因为相似的矩阵有相同的特征多项式，所以
$\det(\lambda E-A) = \det(\lambda E-B)$. 
所以 $$(\lambda -1)(\lambda-x)(\lambda-1)-2(\lambda-1) = (\lambda -1)(\lambda -2)(\lambda -y). $$
消去 $\lambda-1$ 可得
$(\lambda-x)(\lambda-1)-2 = (\lambda -2)(\lambda -y)$. 
即 $\lambda^2-(x+1)\lambda +x-2 = \lambda^2-(2+y)\lambda +2y$. 
比较系数可得 $x+1=2+y, x-2=2y$, 所以 $x=0, y=-1$. 

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  矩阵 $B$ 是对角阵，特征值即为对角线元素 $1, -1, 2$. 所以矩阵 $A$ 也有这些特征值。分别计算相应的特征向量，
可得 $\alpha_1=(-2,6,5)^T, \alpha_2=(0,-2,1)^T, \alpha_3=(0,1,1)^T$, 
故 $P=\left(\alpha_1,\alpha_2,\alpha_3 \right) = \begin{pmatrix}
-2 & 0 & 0 \\
6 & -2 & 1 \\
5 & 1 & 1 \\
\end{pmatrix}
$.

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\end{enumerate}

}

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\item %第6题
\begin{enumerate}
\item  设 $\sigma$ 为 $\mathbb{R}^n$ 上的线性变换。设 $\sigma$ 既是一个正交变换，又是一个对称变换。
证明 $\sigma^2 = \sigma\cdot \sigma=\iota $ 为恒等变换，即对于任意 $\mathbb{R}^n$ 中向量 $\xi$, 我们有 $\sigma(\sigma(\xi))=\xi$. 
\item  设 $A, B$ 都是 $n$ 阶方阵。证明：若 $A$ 与 $B$ 相似, 则 $A$ 与 $B$ 的特征多项式相同。
\end{enumerate}

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{\color{red} 解答：
\begin{enumerate}
\item  设线性变换 $\sigma$ 在某个标准正交基下的矩阵为 $A$. 
因为 $\sigma$ 是对称变换，所以 $A=A^T$. 
因为 $\sigma$ 是正交变换，所以 $A^TA=E$. 
所以有 $A^2=E$, 即 $\sigma^2$ 是恒等变换。

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\item  因为矩阵 $A$ 与 $B$ 相似，所以存在可逆矩阵 $P$, 使得 $B=P^{-1}AP$. 所以 
$$|B-\lambda E| = |P^{-1}AP-\lambda P^{-1}EP| = |P^{-1} (A - \lambda E) P| 
= |P^{-1}|\cdot |A-\lambda E|\cdot  |P|= |A-\lambda E| .$$
其中第三个等号使用了行列式乘积公式 $|AB|=|A|\cdot |B|$, 第四个等号使用了 $$|P^{-1}|\cdot |P|=|P^{-1}P|=|E|=1. $$

\dotfill (\underline{\hspace{0.8cm}分\hspace{0.2cm}})

\end{enumerate}

}

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\end{enumerate}


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